The question is incomplete, here is the complete question:
The rate of certain reaction is given by the following rate law:
![rate=k[H_2]^2[NH_3]](https://tex.z-dn.net/?f=rate%3Dk%5BH_2%5D%5E2%5BNH_3%5D)
At a certain concentration of ![H_2 and [tex]I_2, the initial rate of reaction is 0.120 M/s. What would the initial rate of the reaction be if the concentration of [tex]H_2 were halved.Answer : The initial rate of the reaction will be, 0.03 M/sExplanation :Rate law expression for the reaction:[tex]rate=k[H_2]^2[NH_3]](https://tex.z-dn.net/?f=H_2%20and%20%5Btex%5DI_2%2C%20the%20initial%20rate%20of%20reaction%20is%200.120%20M%2Fs.%20What%20would%20the%20initial%20rate%20of%20the%20reaction%20be%20if%20the%20concentration%20of%20%5Btex%5DH_2%20were%20halved.%3C%2Fp%3E%3Cp%3E%3Cstrong%3EAnswer%20%3A%20The%20initial%20rate%20of%20the%20reaction%20will%20be%2C%200.03%20M%2Fs%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%3Cstrong%3EExplanation%20%3A%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%3Cstrong%3ERate%20law%20expression%20for%20the%20reaction%3A%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%5Btex%5Drate%3Dk%5BH_2%5D%5E2%5BNH_3%5D)
As we are given that:
Initial rate = 0.120 M/s
Expression for rate law for first observation:
....(1)
Expression for rate law for second observation:
....(2)
Dividing 2 by 1, we get:
![\frac{R}{0.120}=\frac{k(\frac{[H_2]}{2})^2[NH_3]}{k[H_2]^2[NH_3]}](https://tex.z-dn.net/?f=%5Cfrac%7BR%7D%7B0.120%7D%3D%5Cfrac%7Bk%28%5Cfrac%7B%5BH_2%5D%7D%7B2%7D%29%5E2%5BNH_3%5D%7D%7Bk%5BH_2%5D%5E2%5BNH_3%5D%7D)


Therefore, the initial rate of the reaction will be, 0.03 M/s
Answer:
1) HBr; 2) NaCCH; 3) O3; 4) H2O
Explanation:
The first step is formation of alkyl halide followed by reaction with sodium acetylide, to form 3-methylbutene, which is then followed by oxidation reaction with O3& H2O to 3-methylbutanal
Answer:
the system has infinitely many solutions.
Explanation:
The system is 2x + y = 1 and 4x + 2y = 2. Solutions to a system are the intersection points. Since these two lines are the same line they intersect everywhere. There are infinitely many solutions.
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Answer:
Chlorine and bromine
Explanation:
I think thats the answer in your question
<u>Answer:</u> The electronic configuration of gallium is written below and number of valence electrons is 3.
<u>Explanation:</u>
Electronic configuration is defined as the representation of electrons around the nucleus of an atom.
Number of electrons in an atom is determined by the atomic number of that atom.
Valence electrons are defined as the electrons present in the outermost shell of an atom.
We are given:
An element Gallium having atomic number as 31.
Number of electrons = 31
Electronic configuration of Gallium is: 
This element has 3 electrons in its outermost shell. So, the number of valence electrons is 3
Hence, the electronic configuration of gallium is written below and number of valence electrons is 3.